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Let $\mathbb Q_{\ge0}$ be the set of all nonnegative rational numbers. I have the following conjecture based on my computation.

4-3-2 Conjecture. Each $r\in\mathbb Q_{\ge0}$ can be written as $x^4+y^3+z^2$ with $x,y,z\in\mathbb Q_{\ge0}$.

As $m/n=(mn^{11})/n^{12}$ for any integers $m\ge0$ and $n>0$, it suffices to consider the 4-3-2 conjecture with $r\in\mathbb N=\{0,1,2,\ldots\}$. For example, \begin{align}2^{12}\times7=&2^4+15^3+159^2, \\4^{12}\times75=&122^4+1007^3+3951^2, \\3^{12}\times1140=&0^4+531^3+21357^2, \\5^{12}\times23710=&217^4+17897^3+232166^2. \end{align} For each $n=1,\ldots,25000$ with $n\not=23710$, I have found a number $m\in\{1,2,3,4\}$ such that $m^{12}n=x^4+y^3+z^2$ for some $x,y,z\in\mathbb N$.

QUESTIONS. Is the 4-3-2 conjecture true? Any way to prove it?

Your comments are welcome!

Let $\mathbb Q_{\ge0}$ be the set of all nonnegative rational numbers. I have the following conjecture based on my computation.

4-3-2 Conjecture. Each $r\in\mathbb Q_{\ge0}$ can be written as $x^4+y^3+z^2$ with $x,y,z\in\mathbb Q_{\ge0}$.

As $m/n=(mn^{11})/n^{12}$ for any integers $m\ge0$ and $n>0$, it suffices to consider the 4-3-2 conjecture with $r\in\mathbb N=\{0,1,2,\ldots\}$. For example, \begin{align}2^{12}\times7=&2^4+15^3+159^2, \\4^{12}\times75=&122^4+1007^3+3951^2, \\3^{12}\times1140=&0^4+531^3+21357^2, \\5^{12}\times23710=&217^4+17897^3+232166^2. \end{align} For each $n=1,\ldots,30000$ with $n\not=23710$, I have found a number $m\in\{1,2,3,4\}$ such that $m^{12}n=x^4+y^3+z^2$ for some $x,y,z\in\mathbb N$.

QUESTIONS. Is the 4-3-2 conjecture true? Any way to prove it?

Your comments are welcome!

Let $\mathbb Q_{\ge0}$ be the set of all nonnegative rational numbers. I have the following conjecture based on my computation.

4-3-2 Conjecture. Each $r\in\mathbb Q_{\ge0}$ can be written as $x^4+y^3+z^2$ with $x,y,z\in\mathbb Q_{\ge0}$.

As $m/n=(mn^{11})/n^{12}$ for any integers $m\ge0$ and $n>0$, it suffices to consider the 4-3-2 conjecture with $r\in\mathbb N=\{0,1,2,\ldots\}$. For example, \begin{align}2^{12}\times7=&2^4+15^3+159^2, \\4^{12}\times75=&122^4+1007^3+3951^2, \\3^{12}\times1140=&0^4+531^3+21357^2. \end{align} For each $n=1,\ldots,20000$, I have found a number $m\in\{1,2,3,4\}$ such that $m^{12}n=x^4+y^3+z^2$ for some $x,y,z\in\mathbb N$.

QUESTIONS. Is the 4-3-2 conjecture true? Any way to prove it?

Your comments are welcome!

Let $\mathbb Q_{\ge0}$ be the set of all nonnegative rational numbers. I have the following conjecture based on my computation.

4-3-2 Conjecture. Each $r\in\mathbb Q_{\ge0}$ can be written as $x^4+y^3+z^2$ with $x,y,z\in\mathbb Q_{\ge0}$.

As $m/n=(mn^{11})/n^{12}$ for any integers $m\ge0$ and $n>0$, it suffices to consider the 4-3-2 conjecture with $r\in\mathbb N=\{0,1,2,\ldots\}$. For example, \begin{align}2^{12}\times7=&2^4+15^3+159^2, \\4^{12}\times75=&122^4+1007^3+3951^2, \\3^{12}\times1140=&0^4+531^3+21357^2, \\5^{12}\times23710=&217^4+17897^3+232166^2. \end{align} For each $n=1,\ldots,25000$ with $n\not=23710$, I have found a number $m\in\{1,2,3,4\}$ such that $m^{12}n=x^4+y^3+z^2$ for some $x,y,z\in\mathbb N$.

QUESTIONS. Is the 4-3-2 conjecture true? Any way to prove it?

Your comments are welcome!

Let $\mathbb Q_{\ge0}$ be the set of all nonnegative rational numbers. I have the following conjecture based on my computation.

4-3-2 Conjecture. Each $r\in\mathbb Q_{\ge0}$ can be written as $x^4+y^3+z^2$ with $x,y,z\in\mathbb Q_{\ge0}$.

As $m/n=(mn^{11})/n^{12}$ for any integers $m\ge0$ and $n>0$, it suffices to consider the 4-3-2 conjecture with $r\in\mathbb N=\{0,1,2,\ldots\}$. For example, \begin{align}2^{12}\times7=&2^4+15^3+159^2, \\4^{12}\times75=&122^4+1007^3+3951^2, \\3^{12}\times1140=&0^4+531^3+21357^2. \end{align} I even guess that for any $n\in\mathbb N$ there is a number $m\in\{1,2,3,4\}$ such that $m^{12}n=x^4+y^3+z^2$ for some $x,y,z\in\mathbb N$. This has been verified for $n\le20000$.

QUESTIONS. Is the 4-3-2 conjecture true? Any way to prove it? Can one find a number $n\in\mathbb N$ such that $m^{12}n\not\in\{x^4+y^3+z^2:\ x,y,z\in\mathbb N\}$ for all $m=1,2,3,4$?

Your comments are welcome!

Let $\mathbb Q_{\ge0}$ be the set of all nonnegative rational numbers. I have the following conjecture based on my computation.

4-3-2 Conjecture. Each $r\in\mathbb Q_{\ge0}$ can be written as $x^4+y^3+z^2$ with $x,y,z\in\mathbb Q_{\ge0}$.

As $m/n=(mn^{11})/n^{12}$ for any integers $m\ge0$ and $n>0$, it suffices to consider the 4-3-2 conjecture with $r\in\mathbb N=\{0,1,2,\ldots\}$. For example, \begin{align}2^{12}\times7=&2^4+15^3+159^2, \\4^{12}\times75=&122^4+1007^3+3951^2, \\3^{12}\times1140=&0^4+531^3+21357^2. \end{align} For each $n=1,\ldots,20000$, I have found a number $m\in\{1,2,3,4\}$ such that $m^{12}n=x^4+y^3+z^2$ for some $x,y,z\in\mathbb N$.

QUESTIONS. Is the 4-3-2 conjecture true? Any way to prove it?

Your comments are welcome!